SOLUTION: Isabelle paid for her $1.75 lunch with 87 coins. If all the coins were nickels and pennies , how many were there of each type of coin

Question 1128491: Isabelle paid for her $1.75 lunch with 87 coins. If all the coins were nickels and pennies , how many were there of each type of coin
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Algebraically....

let n = number of nickels
then (87-n) = number of pennies

The total value of the coins (5 cents for each nickel and 1 cent for each penny) was $1.75 = 175 cents. Write and solve the equation that says that:

You can finish the formal solution.

Informally, using logical reasoning....

(1) If all 87 coins were pennies, that would be only 87 cents; the actual total of 175 cents is 175-87 = 88 cents more than that.
(2) Exchanging a penny for a nickel keeps the same total of 87 coins but increases the total value by 4 cents.
(3) To make up those other 88 cents, the number of nickels needed is 88/4 = 22.

ANSWER: She used 22 nickels and 87-22 = 65 dimes.

CHECK: 22(5)+65(1) = 110+65 = 175
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